Problem: For a function $f$, we are given that $f(3)=-2$ and $f'(3)=4$. What's the equation of the tangent line to the graph of $f$ at $x=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-3=4(x+2)$ (Choice B) B $y-3=-2(x-4)$ (Choice C) C $y-4=-2(x-3)$ (Choice D) D $y+2=4(x-3)$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $f'(3)$ gives the slope of the tangent line to the graph of $f$ where $x=3$. We are given that $f'(3)=4$, so the slope of the tangent line is $4$. Furthermore, we are given that $f(3)=-2$, which means the point of intersection of the tangent line and the graph is $(3,-2)$. To summarize, the tangent line has a slope of $4$ and it passes through the point $(3,-2)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-(-2)&=4(x-3) \\\\ y+2&=4(x-3) \end{aligned}$ The equation is $y+2=4(x-3)$.